Removable Discontinuities

The first way that a function can fail to be continuous at a point a is that

lim

f(x) = L exists (and is finite)

x --> a

but f(a) is not defined or f(a) L. Discontinuities for which the limit of f(x) exists and is finite are called removable discontinuities for reasons explained below.

f(a) is not defined

If f(a) is not defined, the graph has a "hole" at (a, f(a)). This hole can be filled by extending the domain of f(x) to include the point x=a and defining

f(a) =

lim

f(x).

x --> a

This has the effect of removing the discontinuity.

If

lim

f(x) = L but f(a) is not defined

x-->a

then the discontinuity at x=a can be removed by defining f(a)=L.

Graph of (x2 - 1)/(x - 1)

As an example, consider the function g(x) = (x2 - 1)/(x - 1). Then g(x) = x+1 for all real numbers except x=1. Since g(x) and x+1 agree at all points other than the objective,

lim

g(x)

=

lim

x+1 = 2.

x --> 1

x --> 1

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We can "remove" the discontinuity by filling the hole. The domain of g(x) may be extended to include x=1 by declaring that g(1) = 2. This makes g(x) continuous at x=1. Since g(x) is continuous at all other points (as evidenced, for example, by the graph), defining g(x) = 2 turns g into a continuous function.

The limit and the value of the function are different.

If the limit as x approaches a exists and is finite and f(a) is defined but not equal to this limit, then the graph has a hole with a point misplaced above or below the hole. This discontinuity can be removed by re-defining the function value f(a) to be the value of the limit.

If

lim

f(x) = L but f(a) L

x-->a

then the discontinuity at x=a can be removed by re-defining f(a)=L.

As and example, the piecewise function in the second equipment check on the page "Defintion of Continuity" was given by

{

Undefined

Unless 0 < x &lt 1

h(x) =

3

If x=.5

1.5 + 1/(x + .25)

0 < x &lt 1, x.5

Portion of the graph of h(x)

We can remove the discontinuity by re-defining the function so as to fill the hole. In this case we re-define h(.5) = 1.5 + 1/(.75) = 17/6.